2014 AMC 10B Problem 21

Drop perpendiculars from $D$ and $C$ to $\overline{AB}$, meeting it at $D'$ and $C'$. Because the angles at $A$ and $B$ are acute, both feet lie inside $\overline{AB}$, and $D'C' = CD = 21$. Let $x = AD'$ and $y = C'B$; then $x + y = 33 - 21 = 12$.

Take $AD = 10$ and $BC = 14$ (the other assignment is the mirror image and has the same diagonals). With height $h$,
$$x^2 + h^2 = 100, \qquad y^2 + h^2 = 196.$$
Subtracting, $y^2 - x^2 = 96$, so $(y - x)(y + x) = 96$ and $y - x = 8$. Hence $y = 10$, $x = 2$, and $h^2 = 100 - 4 = 96$.

Now compute the diagonals from right triangles with height $h$:

• $AC$: horizontal run $AC' = 33 - 10 = 23$, so $AC^2 = 23^2 + 96 = 529 + 96 = 625$ and $AC = 25$.
• $BD$: horizontal run $BD' = 33 - 2 = 31$, so $BD^2 = 961 + 96 = 1057$, which is larger.

The shorter diagonal is $\boxed{\textbf{(B)}\ 25}$.

Every trapezoid problem with given side lengths reduces to two right triangles sharing the height; the two Pythagorean equations subtract to a difference of squares, which is why the overhangs come out as integers. Once $h^2$ is known, any other segment is one more Pythagorean computation. The "acute angles at $A$ and $B$" clause is what guarantees $x + y = 12$ rather than $|x - y| = 12$.

Slide $\overline{CD}$ along the parallel lines: translating $D$ by $21$ toward $B$ creates triangle with sides $10$, $14$, $12$, whose area by Heron is $\sqrt{18 \cdot 8 \cdot 4 \cdot 6} = 24\sqrt6$; so $h = \frac{2 \cdot 24\sqrt6}{12} = 4\sqrt6$, the same $h^2 = 96$.

• Splitting the overhang evenly (6 and 6) as if the trapezoid were isosceles, which contradicts the unequal legs 10 and 14.
• Computing $BD$ (the longer diagonal, $\sqrt{1057}$) or using the wrong horizontal run, such as $21 + 2 = 23$ for both diagonals.